Question

If x=2-√5/2+√5 and y=2+√5/2-√5 find x^2+y^2​

Answer 1

Answer:

$\left(2-\frac{\sqrt{5}}{2}+\sqrt{5}\right)^2+\left(2+\frac{\sqrt{5}}{2}-\sqrt{5}\right)^2=\frac{21}{2}$

Step-by-step explanation:

$\left(2-\frac{\sqrt{5}}{2}+\sqrt{5}\right)^2+\left(2+\frac{\sqrt{5}}{2}-\sqrt{5}\right)^2$

$\left(2-\frac{\sqrt{5}}{2}+\sqrt{5}\right)^2=\left(2-\frac{\sqrt{5}}{2}+\sqrt{5}\right)\left(2-\frac{\sqrt{5}}{2}+\sqrt{5}\right)$

$=\left(-\frac{\sqrt{5}}{2}+2+\sqrt{5}\right)\left(-\frac{\sqrt{5}}{2}+2+\sqrt{5}\right)+\left(\frac{\sqrt{5}}{2}+2-\sqrt{5}\right)^2$

$\left(2+\frac{\sqrt{5}}{2}-\sqrt{5}\right)^2=\left(2+\frac{\sqrt{5}}{2}-\sqrt{5}\right)\left(2+\frac{\sqrt{5}}{2}-\sqrt{5}\right)$

$=\left(-\frac{\sqrt{5}}{2}+2+\sqrt{5}\right)\left(-\frac{\sqrt{5}}{2}+2+\sqrt{5}\right)+\left(\frac{\sqrt{5}}{2}+2-\sqrt{5}\right)\left(\frac{\sqrt{5}}{2}+2-\sqrt{5}\right)$

$=\frac{5-8\sqrt{5}}{4}+8+\left(2+\frac{\sqrt{5}}{2}-\sqrt{5}\right)\left(2+\frac{\sqrt{5}}{2}-\sqrt{5}\right)$

$=\frac{5-8\sqrt{5}}{4}+8+2\cdot \:2+2\cdot \frac{\sqrt{5}}{2}-2\sqrt{5}+2\cdot \frac{\sqrt{5}}{2}+\frac{\sqrt{5}}{2}\cdot \frac{\sqrt{5}}{2}-\sqrt{5}\frac{\sqrt{5}}{2}-2\sqrt{5}-\sqrt{5}\frac{\sqrt{5}}{2}+\sqrt{5} \sqrt{5}$

$=\frac{21}{2}$